與Servlet映射得到tomcat的錯誤

Question

其實這是我的第一個servlet application.just入門..
下面 是我的代碼：

package newpackage.org;  

import java.io.IOException; 
import java.io.PrintWriter; 
import javax.servlet.ServletException; 

import javax.servlet.http.HttpServlet; 
import javax.servlet.http.HttpServletRequest; 
import javax.servlet.http.HttpServletResponse; 

//@WebServlet(description = "a general servlet", urlPatterns = { 
"/simpleservlet" }) 

public class simpleservlet extends HttpServlet { 

    private static final long serialVersionUID = 1L; 

    /** 
    * @see HttpServletdoGet(HttpServletRequest request, HttpServletResponse 
response) 
    */ 

    protected void doget(HttpServletRequest request, HttpServletResponse 
response) throws ServletException, IOException { 

     // TODO Auto-generated method stub 

    // response.getWriter().append("Served at: ").append(request.getContextPath()); 

    PrintWriter obj= response.getWriter();; 

// String username= request.getParameter("username"); 

    obj.println("again getting back with "); 
    } 
    }` 

XML

<?xml version="1.0" encoding="UTF-8"?> 
    <web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xmlns="http://java.sun.com/xml/ns/javaee" 
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee 

http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0"> 

    <display-name>myproject</display-name> 

    <servlet-mapping> 
<servlet-name>xmlservlet</servlet-name> 
<url-pattern>/002path</url-pattern> 
</servlet-mapping> 
<servlet> 

<servlet-name>xmlservlet</servlet-name> 
<servlet-class>newpackage.org.xmlservlet </servlet-class> 
</servlet> 

<servlet-mapping> 
<servlet-name>simpleservlet</servlet-name> 
<url-pattern>/001path</url-pattern> 
</servlet-mapping> 
<servlet> 

<servlet-name>simpleservlet</servlet-name> 
<servlet-class>newpackage.org.xmlservlet </servlet-class> 
</servlet> 

</web-app> 

錯誤：

HTTP Status 404 - /myproject/servlet/newpackage.org.simpleservlet

type Status report

message /myproject/servlet/newpackage.org.simpleservlet

description The requested resource is not available.

Apache Tomcat/7.0.72

Answer 1

您需要刪除以下在web.xml

<servlet-mapping> 
<servlet-name>simpleservlet</servlet-name> 
<url-pattern>/001path</url-pattern> 
</servlet-mapping> 
<servlet> 
<servlet-name>simpleservlet</servlet-name> 
<servlet-class>newpackage.org.xmlservlet </servlet-class> 
</servlet> 

而在servlet的使用下面的代碼：

@WebServlet(description = "a general servlet", urlPatterns = {"/simpleservlet", "/001path" }) 

Answer 2

<servlet-name>simpleservlet</servlet-name> 
<servlet-class>newpackage.org.xmlservlet </servlet-class> 
</servlet> 

給的，而不是xmlservlet simpleservlet類名

<servlet-name>simpleservlet</servlet-name> 
<servlet-class>newpackage.org.simpleservlet </servlet-class> 
</servlet>